Hello!!!

Could you give me a hint how can we can find it?

is it possible to disprove something, but that a counterexample is impossible to write down? (even abbreviated)

Here's a classic example of that

If $a$, $b$ are irrational, must $a^b$ be irrational?

To disprove that, consider $a=b=\sqrt{2}$. This is either rational or irrational.

If it's rational, it's a counterexample.

If $\sqrt{2}^{\sqrt{2}}$ is instead rational, then $(\sqrt{2}^\sqrt{2})^{\sqrt{2}}=\sqrt{2}^{\sqrt{2}\cdot\sqrt{2}}=\sqrt{2}^2=2$ is rational.

So either $(a,b)=(\sqrt{2},\sqrt{2})$ is an example or $(a,b)=(\sqrt{2}^\sqrt{2},\sqrt{2})$ is an example.

Hence one of these is a counterexample. But which one? @Null

(That's not a perfect example; you can prove which one is the counterexample by other means. But that's a lot harder.)

@Semiclassic: Did you figure out an example for that derivative/limit question?

no, but I wasn't thinking about it

No, they keep you busy in here.

lolyep

on that note, do you have a simple explanation for Mahmoud's question above?

Given the questions that person asked subsequently, I doubt he'll figure this out alone.

Given that he said that he's got an analysis exam tomorrow, I doubt he'll figure it out in time.

Where do I find Mahmoud's question?

If he doesn't understand why $f(x)=\begin{cases} x^2, & x\in\Bbb Q\\ 0, &\text{otherwise}\end{cases}$ has only one derivative at $0$, he's probably in trouble.

starting there

@Evinda: I've told you before that I only taught PDE material once, 30 years ago. You should stop asking me these questions.

It's a bit hazily stated, but I took it as: $f'(x)=(ad-bc)(cx+d)^{-2}$. The factor in the numerator is the determinant of a 2-by-2 matrix. But what does 1-variable calculus have to do with linear algebra?

Hello, chat.

@Semiclassic @Mahmoud: To answer this question in depth, I'd have to use the multivariable chain rule and explain what $\Bbb P^1$ is (over $\Bbb R$ or $\Bbb C$).

hah, I wondered if that's what it was

Oh, Hi @TedShifrin

My own resolution is that it amounts to asking why $ax+by=\lambda(cx+dy)$ for all $x,y$ is equivalent to $ad-bc=0$.

That doesn't totally do it, but it is a heuristic, @Semiclassic.

Right.

And then it comes down to $ax+by=0$, $cx+dy=0$ are dual to the points $[a,b],[c,d]$ which are distinct points in P^1 unless ad=bc.

Something like that.

So the function is constant iff that determinant vanishes, yes.

Right.

I mean, $ax+by=\lambda(cx+dy)\implies a/c=b/d$ is pretty obvious as well.

But I wanted something geometric.

(Assuming you're not dividing by 0 there, @Semiclassic.) At any rate, that is something Mahmoud can understand, but it's not the whole thing.

(Right.)

Gonna have to learn projective geometry when I can.

(Which, yes, @Ted, does mean after Rudin and your book...)

@TedShifrin That is only when the determinant is 0, and the the function becomes just an identity mapping.

It comes back to something I've mentioned before to @Balarka and to @Danu: When you have a mapping $f$ from a manifold to projective space, you can understand its derivative in terms of the derivative of a local smooth lift $\tilde f$ to the associated vector space. That is very geometric. (Visualize a cone and you want to move not along the rulings of the cone.)

not an identity, @Mahmoud, but a constant.

LOL @Fargle. There's some in another one of my books. :P That's what meow is working on.

@TedShifrin Ok, my bad.

I'm proud of being an evil influence on lots of people here. :D

I suppose one could say it like this: If we take the function $f(x)=(ax+b)/(cx+d)$ and lift to P^1, we get $f([x,y])=[ax+by,cx+dy]$

@TedShifrin Mm. I've tried learning it at different points. It's just that my visualization is far too Euclidean and it's hard to divorce myself of that.

not lift to but from on both sides.

Aw come on, @Fargle ... You've seen railroad tracks before.

??

I can't easily do commutative diagrams here, Semiclassic.

Lol, fair enough

You're interpreting a map $\Bbb P^1\to\Bbb P^1$ by lifting to $\Bbb C^2\to\Bbb C^2$. Indeed, a projective transformation is what you get by taking any nonsingular transformation and looking at the induced map on $\Bbb P^1$.

@TedShifrin I mean, sure. It's just for some reason more awkward to me.

Hm.

It's so cool not to have to worry about an exceptional case when lines are parallel, @Fargle.

@Semiclassical 2 is rational? i mean that's a fact or not? or do you mean it is impossible in the proofmaker shoes?

Wait ... What is $\mathbb P^1$ supposed to be ? :o

@null uh

2 is definitely rational.

You'll learn eventually, @Mahmoud. It's $\Bbb R$ together with a point at infinity.

It parametrizes all the lines through the origin in $\Bbb R^2$.

the point is that it's not at all obvious whether $\sqrt{2}^\sqrt{2}$ is rational or not.

Think of slope.

@Semiclassical ah ok

If it is, that's a counterexample to "every irrational power of an irrational is irrational"

@TedShifrin I'm sure--and the elegance of putting a point or line or plane at infinity is neat. It just hurts my intuitive muscles a bit.

Lines through the origin are characterized by their slope $m\in\Bbb R\cup\{\infty\}$.

Well, @Fargle, I'll send you that chapter when you get there. :P

If not, you can use it to construct a counterexample.

Yes lines of the form $y=mx$

@Mahmoud: So it's natural to work with this set. That's what $\Bbb P^1$ is.

Equivalently, they're of the form $ax+by=0$ where $(a,b)\in \mathbb{R}^2$

BTW, @Fargle: Were there other problems you sent me that you want me to read/criticize carefully? I feel like this chapter is silly since you already took Munkres.

That's gonna be confusing, @Semiclassic.

That lets you include stuff like $x=0$. This fits into $y=mx$ if you write $x=\frac1m x$ and take $m=\infty$.

Hey, how can I solve $ \int_1^{2} \dfrac {e^x} {x^2 - 1} $ ?

Is it? hrm

Integrate by parts twice @Maks.

@TedShifrin I suppose that's true. I think it'll get more interesting with Chapter 3, as we didn't do much of the sequence stuff.

Well, $a,b$ were showing up in a totally different context, @Semiclassic.

as I said earlier, maks, need dx.

I corrected it, still integrate by parts ?

hrm.

Oh, you changed the question significantly.

Yes, my bad

That one you probably can't do in elementary terms.

yeah, that's a lot harder.

I think I can prove you can't do it other than numerically.

It wouldn't surprise me.

@TedShifrin want to ask if my example is correct.

consider $f = cos(x)$ and $g = sin(x)$

Actually, I think that integral diverges @maks

It's got a pole at x=1.

then $||f - g||^2 = 1 = ||f + g||^2 \neq 4 = 2(||f||^2 + ||g||^2)$

@Semiclassical Indeed, you can I prove that ?

Why do you say $\|f\pm g\|=1$, Karim?

@Adeek I disagree that that equals 4...

Probably.

@Fargle: I think $2\cdot 2 = 4$ is correct.

we are considering sup though @Fargle ?

Might just be as simple as $e^{x}>e$ for all $x\geq 1$.

But Karim didn't add the two things on the left.

Oh herp.

Karim, I still asked you ^^^^ ...

$||f - g||^2 = (sup_{x \in K} |sin(x) - g(x)|)^2$

i.e. $e^x\geq e$ for $x\geq 1$ implies $\frac{e^x}{x^2-1}\geq \frac{e}{x^2-1}$

Yes, I know the definition, @Karim. But you messed up.

yeah I see

I didn't add them

No, the $1$'s are wrong!!

$||f - g||^2 + ||f + g||^2 = 2 \neq 4 = 2(||f||^2 + ||g||^2)$

smacks Karim for real

And if you can show that $\int_1^2 \frac{e}{x^2-1}\,dx=\infty$ then $\int_1^2 \frac{e^x}{x^2-1}\,dx=\infty$ as well

(not quite the right way to say that but oh well)

yeah it is 4 @TedShifrin yes my mistake.

it is not 1

Oh, @Maks, this was an improper integral question. I didn't even pay attention.

Huh? Karim

What is $4$?

I didn't notice until I tried plugging it into mathematica, cough

Easier just to use partial fractions to split up $1/(x^2-1)$.

Just to get my head on right with the projective stuff, @Ted

$||f + g||^2 = 4$ and $||f - g||^2 = 0$

@TedShifrin

NOOOOO, Karim.

I start with a map $x\mapsto \frac{ax+b}{cx+d}$ from $\Bbb R\to \Bbb R$.

why ?

First of all $\|f-g\|=0$ is totally nonsense.

@TedShifrin like $ \dfrac {e^x} {x^2 - 1} \geq \dfrac {1} {x^2 - 1}$ ?

Second of all, how are you finding $\|f\pm g\|$?

I can get from this a map $[x,1]\mapsto [\frac{ax+b}{cx+d},1]=[ax+b,cx+d]$

@Maks: Have you done partial fractions?

oh I see. I am considering supermum over addition and subtraction not on individually.

Right, @Semiclassic. So it's induced from a map $\Bbb K^2\to\Bbb K^2$.

@TedShifrin Like $ \dfrac {x} {y} = \dfrac {A}{y} + \dfrac {B} {y} $ ?

$\Bbb K^2$?

Like $\dfrac 1{x^2-1} = \dfrac{-1/2}{x+1} + \dfrac{1/2}{x-1}$ or something.

$||f - g||^2 = (sup_{x \in K}|f(x) - g(x)|)^2 = 4$

$\mathbb K^2$ I'm totally lost ..

@Mahmoud That was to me

$\Bbb K = \Bbb R$ or $\Bbb C$.

@TedShifrin Yes I know them, we use them to solve integrals

Ah, fair enough. Whatever the field is.

$\mathbb{K} = \mathbb{R}$

Right, so you can split up the fraction like that and then it's easy to see, @Maks.

Yes, $e^x\ge e^1 = e$. So $\int_1^2\dfrac{e^x}{x^2-1}\,dx \ge \int_1^2\dfrac{e/2}{x-1}\,dx + C$ (and you don't care about $C$).

You mean like use one part of the fraction ? And compare it to the series ?

Ohh ok

right @TedShifrin ?

So we'd say: A linear map $\Bbb K^2\to \Bbb K^2$ induces a projective map $\Bbb{KP}^1\to \Bbb{KP}^1$, which projects to a linear fractional map $\Bbb K\to \Bbb K$?

No, Karim ... Seriously. Slow down and get this right.

No, there's no projects to. A linear map on $\Bbb K^2$ induces a map on $\Bbb KP^1$.

You're just looking at what that final map is on $\Bbb K\subset\Bbb KP^1$ (still mapping to $\Bbb KP^1$, after all, since you may well hit infinity).

Hmm

This proves that my knowledge in Math is very humble ..

Could I say that it restricts to a map on K?

Mahmoud, you're only at the very beginning. So it should be.

In a graphing software I use, all the curves in $\mathbb{E}^3$ with height diverging to $+\infty$ look as though they meet at the same point above the origin. Even $(0,t,e^t)$ for example appears to almost curve backward some to "hit" that point at infinity.

I get that's how the projection is, but it's so weird to look at.

It restricts to give an LFT on $K$, @Semiclassic, but that is still a map to $K\cup\{\infty\} = \Bbb KP^1$.

Hmm.

And when you were taught LFT's, they talked about plugging in infinity, in fact, so really you were doing the map on $\Bbb KP^1$.

True.

Heya @Brody.

@Brody: Try drawing the twisted cubic curve $(t,t^2,t^3)$ by hand :P

So I really shouldn't be thinking of $f(x)=(ax+b)/(cx+d)$ as being a map on K->K at all

If you take liberties, you do, @Semiclassic, but it's not rigorously correct.

It's just the affine representation of such.

But it has a pole at $x=-d/c$.

Hola @Ted

hey guys, anyone with experience with red and black trees ?

Hmm.

Still not entirely satisfied, but perhaps I shouldn't be.

UGA students have experience with all things "red and black." :D

@TedShifrin Don't know what that looks like but I'll actually try some

Where's @Kaj?

Bipartite trees?

It's an interesting curve. You'll see a picture of it in chapter 2 of the multivariable book, in fact.

I'm basically asked build every red and black trees with nodes {1,2,3,4,5,6}

But, a LFT $f:\mathbb{KP}^1\to \mathbb{KP}^1$ acts as $[x,1]\mapsto [ax+b,cx+d]$.

but with no insertion order specified

Sure, @Semiclassic.

@TedShifrin $(t,t^2,t^3)$ just means the usual parametrization in rectangular coords, right?

Yes @Brody.

And this lifts to the linear transformation $[x,y]\mapsto [ax+by,cx+dy]$ on K^2 -> K^2 which induces it.

@TedShifrin A friend of mine did this, is it alright ?

No brackets up there. @Semiclassic

you mean, (x,y) not [x,y]?